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D
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Difficulty:
95%
(hard)
Question Stats:
20% (01:34) correct
80%
(02:14)
wrong
based on 46
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The difference of two positive integers is 10. What are the numbers?
(1) The difference between the LCM and HCF of the two numbers is 310.
(2) The LCM of the two numbers is 63 times their HCF.
(1) The difference between the LCM and HCF of the two numbers is 310.
(2) The LCM of the two numbers is 63 times their HCF.
20 Dec 2020, 17:58
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I think the answer should be D here:
We have the formula \(LCM = \frac{a*b}{GCF}\), we will use \(LCM*GCF = x*(x + 10)\) as we know the numbers have a difference of 10.
Statement 1:
The numbers have a difference of 10 so the GCF is only 10 at most. Furthermore, the GCF can only be 1, 2, 5, or 10 because they have a difference of 10. Plug in LCM = GCF + 310 and we have \((GCF+310)*GCF = x*(x + 10)\). Now let's try these cases:
GCF = 1, we would get 311*1 = x*(x + 10) which has no integer solution, the closest one is 13*23 = 299.
GCF = 2, we would get 312*2 = x*(x + 10) = 624. The closest x we can do is 20*30 = 600 but the next pair is 21*31 = 600+21+31 = 652 so there is no integer solution.
GCF = 5, we get 315*5 = x*(x + 10), this has a solution 35*45 = 315*5.
GCF = 10, we get 320*10 = x*(x + 10) = 3200. The closest x is 50*60 = 3000, the next ones 51*61 or 52*62 are definitely not equal to 3200.
Therefore from statement 1 only GCF = 5 is viable, which in turn gives us the numbers 35 and 45.
Statement 2:
Again plug in \(LCM = GCF*63\) and we would have \(63 * GCF^2 = x*(x + 10)\). If we want to split \(63*GCF^2\) into the original numbers, each number must have a copy of the GCF so we have to split the 63 into two factors, then we have only the following cases:
\((1*GCF)* (63GCF)\), \((3*GCF) *(21*GCF)\), \((7*GCF)*(9*GCF)\).
Furthermore, the two numbers must have a difference of 10. The first case gives a difference of 62*GCF which cannot be 10. The 2nd case gives a difference of 18*GCF which cannot be 10. Only the 3rd case gives a difference of 2*GCF and we have GCF = 5, with the two numbers being 35 and 45. Then only one case suffices and this statement is sufficient.
Ans: D
We have the formula \(LCM = \frac{a*b}{GCF}\), we will use \(LCM*GCF = x*(x + 10)\) as we know the numbers have a difference of 10.
Statement 1:
The numbers have a difference of 10 so the GCF is only 10 at most. Furthermore, the GCF can only be 1, 2, 5, or 10 because they have a difference of 10. Plug in LCM = GCF + 310 and we have \((GCF+310)*GCF = x*(x + 10)\). Now let's try these cases:
GCF = 1, we would get 311*1 = x*(x + 10) which has no integer solution, the closest one is 13*23 = 299.
GCF = 2, we would get 312*2 = x*(x + 10) = 624. The closest x we can do is 20*30 = 600 but the next pair is 21*31 = 600+21+31 = 652 so there is no integer solution.
GCF = 5, we get 315*5 = x*(x + 10), this has a solution 35*45 = 315*5.
GCF = 10, we get 320*10 = x*(x + 10) = 3200. The closest x is 50*60 = 3000, the next ones 51*61 or 52*62 are definitely not equal to 3200.
Therefore from statement 1 only GCF = 5 is viable, which in turn gives us the numbers 35 and 45.
Statement 2:
Again plug in \(LCM = GCF*63\) and we would have \(63 * GCF^2 = x*(x + 10)\). If we want to split \(63*GCF^2\) into the original numbers, each number must have a copy of the GCF so we have to split the 63 into two factors, then we have only the following cases:
\((1*GCF)* (63GCF)\), \((3*GCF) *(21*GCF)\), \((7*GCF)*(9*GCF)\).
Furthermore, the two numbers must have a difference of 10. The first case gives a difference of 62*GCF which cannot be 10. The 2nd case gives a difference of 18*GCF which cannot be 10. Only the 3rd case gives a difference of 2*GCF and we have GCF = 5, with the two numbers being 35 and 45. Then only one case suffices and this statement is sufficient.
Ans: D
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